- What are all the individual results that could occur on one spin? - An event can contain one or more outcomes from the sample space.

1. The possible individual outcomes are R, G, and B, so \(S=\{R,G,B\}\). 2. An outcome is one possible result. The set \(\{G,B\}\) contains two outcomes, so it is an event.

a) \(S=\{R,G,B\}\). b) \(\{G,B\}\) is an event because it contains two possible outcomes.

- Which different numbers could you record after a single spin? - Does a number appearing twice create two different recorded results?

1. The distinct displayed numbers are \(1,2,3,4\). 2. A sample space lists different possible recorded results, not one entry for every sector. Therefore, \(S=\{1,2,3,4\}\).

\(S=\{1,2,3,4\}\). Repeated labels represent the same recorded outcome, so each possible number appears once.

- Pair each possible coin result with every possible number. - Because the two stages record different kinds of results, keep their order fixed.

1. The coin result can be H or T, and each can be paired with any of the three numbers. 2. The six ordered outcomes are \((H,1)\), \((H,2)\), \((H,3)\), \((T,1)\), \((T,2)\), and \((T,3)\).

\(S=\{(H,1),(H,2),(H,3),(T,1),(T,2),(T,3)\}\).

- A sample space names distinct results, not individual sectors. - To find a probability, compare the favorable sectors with all equally sized sectors.

1. The distinct results are A, B, and C, so the sample space is \(S=\{A,B,C\}\). 2. A appears on three of the five equal sections, so \(P(A)=\frac{3}{5}\).

a) \(S=\{A,B,C\}\) b) \(P(A)=\frac{3}{5}\)

- For one card, which numbers could the spinner show? - Repeat for the other card, keeping card result first.

1. Pair each card with each possible spinner result: \((X,1),(X,2),(X,3),(Y,1),(Y,2),(Y,3)\). 2. The outcomes \((X,2)\) and \((Y,2)\) differ because they record different cards.

\(S=\{(X,1),(X,2),(X,3),(Y,1),(Y,2),(Y,3)\}\). The card results differ.

- List the favorable outcomes for each event before adding probabilities. - Remember that \(1\) is not prime. - Identify the perfect squares from \(1\) through \(8\). - For an “or” event, include every outcome satisfying at least one condition, without double-counting overlaps.

1. Event \(A\) contains \(1\) and \(4\), so \(P(A)=0.05+0.20=0.25\). 2. Event \(B\) contains \(5\), \(6\), \(7\), and \(8\), so \(P(B)=0.08+0.12+0.15+0.15=0.50\). 3. Event \(C\) contains the primes \(2\), \(3\), \(5\), and \(7\), so \(P(C)=0.15+0.10+0.08+0.15=0.48\). 4. The odd outcomes are \(1\), \(3\), \(5\), and \(7\), and the outcomes divisible by \(4\) are \(4\) and \(8\). These sets are disjoint, so \(P(D)=0.05+0.10+0.08+0.15+0.20+0.15=0.73\).

\(A\): \(P(A)=0.25\) \(B\): \(P(B)=0.50\) \(C\): \(P(C)=0.48\) \(D\): \(P(D)=0.73\)

- Which pairs of numbers from \(1\) to \(6\) are exactly two apart? - Does reversing the results give the same ordered outcome or a different one? - Check the possible first-roll results so no valid pair is missed.

1. An outcome is an ordered pair \((a,b)\), where both entries are from \(1\) to \(6\). 2. The two results must be two apart, so the smaller result can be \(1,2,3\), or \(4\). Both orders are possible for each choice. 3. The event is \(E=\{(1,3),(2,4),(3,1),(3,5),(4,2),(4,6),(5,3),(6,4)\}\).

\(E=\{(1,3),(2,4),(3,1),(3,5),(4,2),(4,6),(5,3),(6,4)\}\).

- What numbers of heads and tails add to three and differ by one? - Keep track of the order of the tosses, not just how many heads occurred. - Check every possible position for the single head.

1. A three-toss result has three positions. One more tail than head means one head and two tails. 2. The head can appear first, second, or third. The three outcomes are \((H,T,T)\), \((T,H,T)\), and \((T,T,H)\). 3. Therefore, the event is \(E=\{(H,T,T),(T,H,T),(T,T,H)\}\).

\(E=\{(H,T,T),(T,H,T),(T,T,H)\}\).

- What two conditions must every finite probability model satisfy? - Check the total probability before changing any value. - Only one probability is allowed to change.

1. Every listed probability is between \(0\) and \(1\), but their sum is \(0.25+0.35+0.20+0.25=1.05\), so the model is not valid. 2. The first three probabilities sum to \(0.80\). To make the total \(1\), \(P(D)\) must be \(1-0.80=0.20\).

No. The probabilities sum to \(1.05\). The model becomes valid if \(P(D)=0.20\).

- Ask whether heads followed by tails is the same ordered result as tails followed by heads. - After completing the sample space, select only outcomes with one H and one T.

1. The order of the two tosses matters. The outcome TH is possible but is missing, so the complete sample space is \(\{HH,HT,TH,TT\}\). 2. Exactly one head occurs in HT and TH, so \(E=\{HT,TH\}\).

a) The missing outcome is \(TH\), so \(S=\{HH,HT,TH,TT\}\). b) \(E=\{HT,TH\}\).

- Are the two labels carried by the same number of tickets? - Identify what is equally likely: a label or an individual ticket.

1. There are two distinct labels, but eight equally likely tickets. 2. Two tickets have the label YES. Thus \(P(\text{YES})=\frac{2}{8}=\frac{1}{4}\). 3. Ella is incorrect because the labels are not equally likely.

No. \(P(\text{YES})=\frac{1}{4}\).

- What must the probabilities of all outcomes add to? - Write the relationship between the two unknown probabilities using one variable. - To find the probability of an event, add the probabilities of its outcomes.

1. Let \(P(x_4)=p\). Then \(P(x_3)=3p\). 2. Use the fact that all outcome probabilities add to \(1\): \(0.12+0.28+3p+p=1\). 3. Solve: \(0.40+4p=1\), so \(4p=0.60\) and \(p=0.15\). Therefore, \(P(x_4)=0.15\) and \(P(x_3)=0.45\). 4. Add the probabilities of the outcomes in \(E\): \(P(E)=0.12+0.45=0.57\).

\(P(x_3)=0.45\), \(P(x_4)=0.15\), and \(P(E)=0.57\).

- The probabilities of all outcomes must add to \(1\). - Which outcome is not in event \(A\)? Use a complement. - A set with three elements has \(2^3\) subsets. - An event is a subset of the sample space. Add the probabilities of the outcomes it contains.

1. Since \(A^c=\{\omega_3\}\), \(P(\omega_3)=1-P(A)=1-0.55=0.45\). 2. Since \(B^c=\{\omega_1\}\), \(P(\omega_1)=1-P(B)=1-0.75=0.25\). 3. The outcome probabilities add to \(1\), so \(P(\omega_2)=1-0.25-0.45=0.30\). 4. Add the appropriate outcome probabilities for each subset of \(S\): \(P(\emptyset)=0\), \(P(\{\omega_1\})=0.25\), \(P(\{\omega_2\})=0.30\), \(P(\{\omega_3\})=0.45\), \(P(\{\omega_1,\omega_2\})=0.55\), \(P(\{\omega_1,\omega_3\})=0.70\), \(P(\{\omega_2,\omega_3\})=0.75\), and \(P(S)=1\).

Individual outcomes: \(P(\omega_1)=0.25\), \(P(\omega_2)=0.30\), \(P(\omega_3)=0.45\) Events: \(P(\emptyset)=0\), \(P(\{\omega_1\})=0.25\), \(P(\{\omega_2\})=0.30\), \(P(\{\omega_3\})=0.45\), \(P(\{\omega_1,\omega_2\})=0.55\), \(P(\{\omega_1,\omega_3\})=0.70\), \(P(\{\omega_2,\omega_3\})=0.75\), \(P(S)=1\)

- Represent the ordered outcomes in a grid or organized list. - For part b), count every outcome with a \(7\) on the first die, then count the outcomes with a \(7\) on the second die that have not already been counted. - Identify the powers of \(2\) available on each die.

1. There are \(8\cdot12=96\) equally likely ordered outcomes. 2. A sum of \(10\) occurs for \((1, 9),(2, 8),(3, 7),(4, 6),(5, 5),(6, 4),(7, 3),(8, 2)\), so the probability is \(\frac{8}{96}=\frac{1}{12}\). 3. For at least one \(7\), there are \(12\) outcomes with the eight-sided die showing \(7\). There are \(7\) additional outcomes with the twelve-sided die showing \(7\) and the eight-sided die showing a value other than \(7\). Thus there are \(12+7=19\) favorable ordered outcomes, so the probability is \(\frac{19}{96}\). 4. A product is a power of \(2\) only when each factor is a power of \(2\). Each die can show \(1,2,4,8\), giving \(4\cdot4=16\) favorable outcomes. Thus, the probability is \(\frac{16}{96}=\frac{1}{6}\).

a) \(\frac{1}{12}\) b) \(\frac{19}{96}\) c) \(\frac{1}{6}\)

- List the ordered pairs that give each prime sum. - For the product event, how does the first roll affect which second rolls can work? - Check how many ordered pairs produce the smallest sum and how many produce a middle sum.

1. The \(36\) ordered pairs are equally likely. 2. Prime sums are \(2,3,5,7,11\). They occur in \(1+2+4+6+2=15\) pairs, so \(P(E_1)=\frac{15}{36}=\frac{5}{12}\). 3. To make the product a multiple of \(4\), a first roll of \(4\) works with all six second rolls. A first roll of \(2\) or \(6\) works with the three even second rolls, giving \(2\cdot3=6\) more pairs. A first roll of \(1\), \(3\), or \(5\) works only if the second roll is \(4\), giving three more pairs. Thus \(P(E_2)=\frac{6+6+3}{36}=\frac{5}{12}\). 4. Jisoo's claim is false. Ordered pairs are equally likely, but sums are not. A sum of \(2\) occurs in one pair, while a sum of \(7\) occurs in six pairs.

1. \(P(E_1)=\frac{5}{12}\) and \(P(E_2)=\frac{5}{12}\). 2. The claim is false because different sums occur in different numbers of equally likely ordered pairs.

- Treat the two dice as distinguishable, so outcomes are ordered pairs. - Start with the smaller result to list the pairs with difference \(2\). What happens if you reverse the order? - In part 2, count the cases with neither die showing \(7\) nor \(8\).

1. There are \(8\cdot8=64\) equally likely ordered pairs. 2. The pairs with the smaller result first and difference \(2\) are \((1, 3),(2, 4),(3, 5),(4, 6),(5, 7),(6, 8)\). Each has a different reversed pair, so there are \(6\cdot2=12\) favorable outcomes and the probability is \(\frac{12}{64}=\frac{3}{16}\). 3. The complement of at least one result being \(7\) or \(8\) is that both dice show a number from \(1\) to \(6\). There are \(6\cdot6=36\) such outcomes, so the probability is \(1-\frac{36}{64}=\frac{7}{16}\).

1. \(\frac{3}{16}\). 2. \(\frac{7}{16}\).

- Make a grid of ordered results for the two rolls; different sums do not occur equally often. - Translate each condition into a list of allowed sum values before counting pairs. - Which possible sums are prime? Remember that the sums themselves are not equally likely.

1. There are \(6\cdot6=36\) equally likely ordered outcomes. 2. In a), the sums are \(4,5,6\); they occur in \(3+4+5=12\) ways. The probability is \(\frac{12}{36}=\frac{1}{3}\). 3. In b), the prime sums are \(2,3,5,7,11\); they occur in \(1+2+4+6+2=15\) ways. The probability is \(\frac{15}{36}=\frac{5}{12}\). 4. In c), the sums are \(2,3,11,12\); they occur in \(1+2+2+1=6\) ways. The probability is \(\frac{6}{36}=\frac{1}{6}\).

a) Sums \(4,5,6\); probability \(\frac{1}{3}\). b) Sums \(2,3,5,7,11\); probability \(\frac{5}{12}\). c) Sums \(2,3,11,12\); probability \(\frac{1}{6}\).

- How many ordered results can two six-sided dice produce? - How can a grid show the different numbers of pairs that give each sum? - In c), might it be easier to count the excluded sums first?

1. There are \(6\cdot6=36\) equally likely ordered pairs. 2. For a), sums \(5,6,7,8\) occur in \(4+5+6+5=20\) pairs, so the probability is \(\frac{20}{36}=\frac{5}{9}\). 3. For b), the odd sums above \(8\) are \(9,11\). They occur in \(4+2=6\) pairs, so the probability is \(\frac{6}{36}=\frac{1}{6}\). 4. For c), exclude sums \(7\) and \(11\), which occur in \(6+2=8\) pairs. The other possible sums are \(2,3,4,5,6,8,9,10,12\). The probability is \(1-\frac{8}{36}=\frac{7}{9}\).

a) Sums \(5,6,7,8\); probability \(\frac{5}{9}\). b) Sums \(9,11\); probability \(\frac{1}{6}\). c) Sums \(2,3,4,5,6,8,9,10,12\); probability \(\frac{7}{9}\).

- Consider the ordered outcomes, not just the possible sums. - Which pairs of factors make a perfect square? How could you group numbers that have the same square-free part? - For a very large difference, which values must be near the ends of the range? - For prime sums, count the pairs for each allowable sum.

1. There are \(12\cdot12=144\) equally likely ordered pairs. 2. Ten ordered pairs have sum \(15\), so the probability is \(\frac{10}{144}=\frac{5}{72}\). 3. A product is a perfect square when the two factors have the same prime factors raised to odd powers. From \(1\) to \(12\), the matching groups are \(\{1,4,9\}\), \(\{2,8\}\), \(\{3,12\}\), and the single-number groups \(\{5\},\{6\},\{7\},\{10\},\{11\}\). Each ordered pair from within one group works. The count is \(3^2+2^2+2^2+1+1+1+1+1=22\), so the probability is \(\frac{22}{144}=\frac{11}{72}\). 4. Differences of at least \(10\) occur for \((1,11),(1,12),(2,12)\) and their reverses. Thus the probability is \(\frac{6}{144}=\frac{1}{24}\). 5. The possible prime sums are \(2,3,5,7,11,13,17,19,23\). They occur in \(1+2+4+6+10+12+8+6+2=51\) ordered pairs, giving \(\frac{51}{144}=\frac{17}{48}\).

1. \(\frac{5}{72}\). 2. \(\frac{11}{72}\). 3. \(\frac{1}{24}\). 4. \(\frac{17}{48}\).